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\begin{center}
2011 Missouri Collegiate Mathematics Competition \\ Session I
\end{center}


\noindent 1.  Consider the parabola $x^2=4\,p\,y$, with vertex at the origin, focus at
$F(0,p)$, and directrix $y= -p$.  Let $P(x,y)$ be a point on the parabola and let
$R(x,-p)$ be the intersection of the vertical line through $P$ with the directrix.
Determine $P$ so that the triangle $PFR$ is equilateral.
\ \\ \ \\
Solution.
\ \\ \ \\
\noindent By the definition of a parabola,
$$dist(F,P) = dist(P,R) = \frac{x^2+4p^2}{4p} .$$
$dist(F,R) = \sqrt{4p^2+x^2}$, so we need
$$\frac{x^2+4p^2}{4p} =  \sqrt{4p^2+x^2}$$
which is equivalent to
$$\sqrt{4p^2+x^2} = 4p$$
making $x=2\sqrt{3}p$ and $y=3p$.
\ \\ \ \\
\noindent 2. Prove that the inequality $\sin x + \arcsin x > 2x$ holds for all values of $x$ such that
$0 < x \le 1$.
\ \\ \ \\
Solution I.
\ \\ \ \\
\noindent The Maclaurin series for $\sin x$ and $\arcsin x$ are given by

\begin{eqnarray*}
\sin x & = & x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots \\
\arcsin x & = & x + \left( \frac{1}{2} \right) \cdot \frac{x^3}{3} + \left( \frac{1}{2} \cdot \frac{3}{4}
\right) \cdot \frac{x^5}{5} + \left( \frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \right) \cdot
\frac{x^7}{7}
+ \cdots .\\
\end{eqnarray*}

\noindent Therefore, for each $0 < x \le 1$, there exists a $\theta_1$ and
$\theta_2$ between 0 and x such that

\begin{eqnarray*}
\sin x & = & x - \frac{x^3}{3!} + \frac{\theta_1^5}{5!} \\
\arcsin x & = & x + \left( \frac{1}{2} \right) \cdot \frac{x^3}{3} + \left( \frac{1}{2} \cdot \frac{3}{4}
\right) \cdot \frac{\theta_2^5}{5} .\\
\end{eqnarray*}

\noindent Therefore,
$$\sin x + \arcsin x = 2x + \frac{\theta_1^5}{5!} + \left( \frac{1}{2} \cdot \frac{3}{4} \right) \cdot
\frac{\theta_2^5}{5} > 2x .$$
\ \\ \ \\
Solution II.
\ \\ \ \\
\noindent Let $f(x)=\sin{x} + \arcsin{x} $ and note that $f(0)=0$.  Then
$\ds{f^{\,\prime}(x) = \cos{x} + \frac{1}{\sqrt{1-x^2}}}$ and $f^{\,\prime}(0) =2$,
making $y=2x$ the tangent line to the graph of $f$ at the origin.
Because $\ds{f^{\,\prime\prime}(x) = -\sin{x} + \frac{x}{(1-x^2)^{3/2}}}$, and
$\ds{\frac{x}{(1-x^2)^{3/2}} > x > \sin{x}}$ for $0 < x < 1$, the graph of $f$
is concave up on $(0,1)$, and thus above the tangent line at $(0,0)$, giving the required inequality.
\ \\ \ \\
\noindent This problem is from \textit{How do you Figure? Math Challenges, M283},
Quantum, \textbf{10.3} (Jan/Feb 2000), 21, 46--47.
\ \\ \ \\
\noindent 3. For what non-negative integers $n$ does there exist a
polynomial $P_n$ of degree $n$ with integer coefficients satisfying
$P_n(k)=2^k$ for all integers $k$, $0 \leq k \leq n$? Find all such
polynomials.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Suppose that $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_1x+a_0$. First assume that $n>1$. Since
$P(0)=1$, $a_0=1$. Then, since $P(2)=4$, we have
$2^na_n+2^{n-1}a_{n-1}+\cdots +2a_1=3$.  Since the parity of
the left-hand-side and right-hand-side are
different, no such polynomial exists. Now assume that $n \leq 1$.
When $n=0$, the polynomial $P_0(x)=1$ does the trick. When
$n=1$, the polynomial $P_1(x)=x+1$ satisfies both $P_1(0)=1$ and
$P_1(1)=2$. Moreover, since $P_n(0)=1$ when $n=0,1$ and $P_1(1)=2$, $P_0$ and $P_1$ are the only such
polynomials.
\ \\ \ \\
\noindent 4. Let $a$ and $b$ be real numbers with $b\neq0$ and let $C\left( \mathbb{R}\right)$
denote the set of continuous functions from the reals to the reals. Define $T:C\left(
\mathbb{R}\right)  \rightarrow C\left(  \mathbb{R}\right)  $ by $\left(
Tf\right)  \left(  x\right)  =ax+b\int_{0}^{x}f\left(  t\right)  dt$. Find all
fixed points of $T^{2}$, where $T^{2}$ denotes the composition of $T$ with
itself.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent A function $f$ is a fixed point of $T^{2}$ if and only if for all real
$x$
\begin{align*}
f\left(  x\right)   & =\left(  T^{2}\right)  \left(  f\right)  \left(
x\right) , \\
f\left(  x\right)   & =ax+b\int_{0}^{x}\left(  Tf\right)  \left(  t\right)
dt , \\
f\left(  x\right)   & =ax+b\int_{0}^{x}\left\{  at+b\int_{0}^{t}f\left(
s\right)  ds\right\}  dt.
\end{align*}
Since $f$ is continuous, the right-hand side is a differentiable function of
$x$; thus the left-hand side is also. Differentiating gives%
\[
f^{\prime}\left(  x\right)  =a+abx+b^{2}\int_{0}^{x}f\left(  s\right)  ds.
\]
Again, both sides are differentiable. Differentiating gives%
\[
f^{\prime\prime}\left(  x\right)  =ab+b^{2}f\left(  x\right)  .
\]
This is a constant coefficient differential equation; the general form of the
solution is $f\left(  x\right)  =c_{1}e^{bx}+c_{2}e^{-bx}+k$, for some
constants $c_{1}$, $c_{2}$, and $k$. Substituting in the differential equation
gives $k=-\dfrac{a}{b}$. From the above, the initial conditions are $f\left(
0\right)  =0$ and $f^{\prime}\left(  0\right)  =a$. Since $f^{\prime}\left(
x\right)  =bc_{1}e^{bx}-bc_{2}e^{-bx}$, we have%
\[
\left\{
\begin{array}
[c]{c}%
c_{1}+c_{2}-\dfrac{a}{b}=0 ;\\
b\left(  c_{1}-c_{2}\right)  =a ,
\end{array}
\right.
\]
so that $c_{1}=\dfrac{a}{b}$ and $c_{2}=0$. Thus, the unique fixed point is%
\[
f\left(  x\right)  =\dfrac{a}{b}\left(  e^{bx}-1\right)  .
\]
\ \\ \ \\
\noindent 5. Let the function $F$ be given by
$$F(x) = e^{-x} - \left(1-\frac{x}{n}\right)^n.$$
Show that, for $n\ge 2$,
$$0\le F(x) \le \frac{e^{-1}}{n} \qquad \hbox{on} \qquad [0,n].$$
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent The function $F$ is continuous and differentiable on $[0,n]$, and so has both a
maximum and a minimum there,
either at an end point or at a critical number.  We have $F(0)=0$, $F(n)=e^{-n}$, and
$$F^{\,\prime}(x) = -e^{-x}+\left(1-\frac{x}{n}\right)^{n-1}.$$
To show $F(x)\ge 0$, we need to show
$$\left(1-\frac{x}{n}\right)^n \le e^{-x}\qquad \hbox{on} \qquad [0,n],$$
or
$$\left(1-\frac{x}{n}\right)^n \le \left(e^{-\frac{x}{n}}\right)^n \qquad \hbox{on} \qquad [0,n],$$
or
$$0 \le \left(1-\frac{x}{n}\right) \le e^{-\frac{x}{n}} \qquad \hbox{on} \qquad [0,n].$$
This is equivalent to showing
$$0 \le \left(1-t \right) \le e^{-t} \qquad \hbox{on} \qquad [0,1],$$
which is clear, because $y=1-t$ is the tangent line to $y=e^{-t}$ at the point $(0,1)$, and the
graph of $y=e^{-t}$ is concave up everywhere and thus is above all its tangent lines.
So the minimum is $0$, which is attained at $x=0$.

Because $F^{\,\prime}(n)=-e^{-n} <0$, the maximum does not occur at $x=n$ and
so must be at a point $x_0$ in $(0,n)$ where $F^{\,\prime}(x_0)=0$.  This means
$$ e^{-x_0} = \left(1-\frac{x_0}{n}\right)^{n-1}$$
and
\begin{eqnarray*}
F(x_0) &=& e^{-x_0} - \left(1 - \frac{x_0}{n}\right)^n \\
    &=& e^{-x_0} - \left(1 - \frac{x_0}{n}\right)^{n-1}\left(1-\frac{x_0}{n}\right) \\
    &=& e^{-x_0} - e^{-x_0}\left(1-\frac{x_0}{n}\right) = \frac{x_0e^{-x_0}}{n}.
\end{eqnarray*}
Now let $g(x) = x\,e^{-x}$.  Basic calculus shows that $g$ has a maximum of $e^{-1}$ at $x=1$, and so
$$F(x_0) \le \frac{e^{-1}}{n}.$$
This problem is a lemma in the article by Levrie and Daems, Evaluating the probability integral
using Wallis' product formula for $\pi$, {\it The American Mathematical Monthly} 116, (2009) 538--541.
Later in the paper, $x$ is replaced by $x^2$ and $n \to \infty$.
\vfill\eject
\begin{center}
2011 Missouri Collegiate Mathematics Competition \\ Session II
\end{center}

\noindent 1. For $0<r<1$, let $n_{r}$ denote the line that is normal to the curve
$y=x^{r}$ at the point $\left(  1,1\right)  $, and let $S_{r}$ denote the
region in the first quadrant of the $xy$-plane bounded by the $x$-axis, the
curve $y=x^{r}$, and the line $n_{r}$. Find the value of $r$ that minimizes
the area of $S_{r}$.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent From $y'=rx^{r-1}$, the slope of the curve $y=x^{r}$ at $\left(
1,1\right)  $ is $r$; hence the equation of line $n_{r}$ is
\[
y-1=-\dfrac{1}{r}\left(  x-1\right)  ,
\]
or equivalently,
\[
x=r\left(  1-y\right)  +1.
\]
The area of $S_{r}$ is given by%
\begin{align*}
A\left(  r\right)   & =\int_{0}^{1}\left[  r\left(  1-y\right)  +1-y^{1/r}%
\right]  dy\\
& =\left(  ry-\dfrac{1}{2}ry^{2}+y-\dfrac{1}{\frac{1}{r}+1}y^{\frac{1}{r}%
+1}\right)  \bigg\vert _{0}^{1}\\
& =r-\dfrac{1}{2}r+1-\dfrac{r}{r+1}\\
& =\dfrac{1}{2}r+\dfrac{1}{r+1}.
\end{align*}
The only critical value of $A\left(  r\right)  $ for $0<r<1$ is the unique
solution in that interval to $\dfrac{1}{2}-\dfrac{1}{\left(  r+1\right)  ^{2}%
}=0$, namely, $r=\sqrt{2}-1$. Since $A^{\prime\prime}\left(  r\right)
=\dfrac{2}{\left(  r+1\right)  ^{3}}$, $A^{\prime\prime}\left(  \sqrt
{2}-1\right)  >0$, implying that the unique local minimum of $A\left(
r\right)  $ is at $r=\sqrt{2}-1$. We have $A\left(  \sqrt{2}-1\right)
=\sqrt{2}-\dfrac{1}{2}$. Since $\lim\limits_{r\rightarrow0^{+}}A\left(
r\right)  =1$ and $\lim\limits_{r\rightarrow1^{-}}A\left(  r\right)  =+\infty
$, $A\left(  r\right)  $ has a global minimum at $r=\sqrt{2}-1$.
\ \\ \ \\
\noindent 2. A rectangular strip is
divided into $n$ bands of equal width  as in the following figure. \\ \ \\
\centerline{\begin{tabular}{|c|c|c|c|c|c|c|c|c|}
\hline 1&2&\phantom{2}&\phantom{2}&\phantom{2}&\phantom{2}&\phantom{2}&n \\ \hline \end{tabular}}
\ \\
Each band is colored by one of $m$
colors. Two patterns are considered identical if one is a (left-to-right) mirror reflection of the other.
Determine the number of distinct patterns with $m$ colors and $n$ bands.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent If a color assignment is not altered by
a mirror reflection, then for any $i$, bands $i$ and $n+1-i$ must
have the same color. If $n$ is even, we can assign arbitrary colors
to bands $1, 2, \ldots , n/2$, in which case there are $m^{n/2}$
color schemes invariant under a mirror reflection. If $n$ is odd,
then the bands $1, 2, \ldots , (n+1)/2$ can be given an arbitrary
coloring scheme and hence there are $m^{(n+1)/2}$ color schemes
invariant under mirror reflection.

Since there are a total of $m^n$ coloring schemes, and since we can put in pairs,
the schemes which are not invariant under mirror reflection, there are a total of
$m^{n/2}+\frac{1}{2}(m^n-m^{n/2})=\frac{1}{2}(m^n+m^{n/2})$ distinct patterns when $n$ is
even and  $m^{(n+1)/2}+\frac{1}{2}(m^n-m^{(n+1)/2})=\frac{1}{2}(m^n+m^{(n+1)/2})$ distinct patterns
when $n$ is odd.
\ \\ \ \\
\noindent 3. Let $M$ be an arbitrary $3 \times 3$ matrix, each of whose entries $m_{ij}$ is drawn
from the set $\{ -1,
1 \}$, and let $\det M$ be the determinant of $M$.  Prove that $\vert \det M \vert \le 4$.
\ \\ \ \\
\noindent Solution I.
\ \\ \ \\
\noindent A $2 \times 2$ matrix with entries $\pm 1$ necessarily has determinant $0$ or $\pm 2$.
Expansion of the
$3 \times 3$ $\det M$ in minors of the elements in the first row gives
\begin{equation}
\det M = m_{11} \left\vert \begin{matrix} m_{22} & m_{23} \\ m_{32} & m_{33} \end{matrix} \right\vert
- m_{12} \left\vert \begin{matrix} m_{21} & m_{23} \\ m_{31} & m_{33} \end{matrix} \right\vert
+ m_{13} \left\vert \begin{matrix} m_{21} & m_{22} \\ m_{31} & m_{33} \end{matrix} \right\vert ,
\end{equation}
so the left-hand side must be even.  Also, it is apparent from (1) that $\vert \det M \vert$ cannot exceed
$6$.  It can be $6$ only if none of the minors is $0$.  But in the $2 \times 3$ matrix
$$\left( \begin{matrix} m_{21} & m_{22} & m_{23} \\ m_{31} & m_{32} & m_{33} \end{matrix} \right) ,$$
where the entries are $1$'s and $-1$'s, it is always possible to find a $2 \times 2$ submatrix (by
striking out some particular column) whose determinant is $0$, regardless of the distribution of
$1$'s and $-1$'s.  Thus, at least one of the terms in (1) must vanish, and we have $\vert \det M \vert
\le 4$.
\ \\ \ \\
\noindent Solution II.
\ \\ \ \\
\noindent Hadamard's Determinantal Inequality says that for any $n \times n$ real matrix $M_n$, $n>2$,
one has
$$\vert \det M_n \vert \le \prod_{j=1}^n \left\{ \sum_{i=1}^n m_{ij}^2 \right\} ^{1/2} .$$
For matrices of the sort considered here, the right-hand side of the inequality is $3 \sqrt 3 \approx 5.2$.
Again, since $\det M_3$ must be even, then $\vert \det M_3 \vert \le 4$ follows.
\ \\ \ \\
\noindent 4. Calculate
$$\int_0^2 \left( \sqrt {1+x^3} + \root 3 \of {x^2 + 2x} \right)\, dx .$$
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent Consider the rectangle with the vertices $O(0,0)$, $A(2,0)$, $B(2,3)$, and $C(0,3)$
on the coordinate plane.  The graph of the function $y=\sqrt {1+x^3}$ passes through the
points $(0,1)$ and $(2,3)$ and partitions our rectangle into two parts.  The area under the
graph is
$$\int_0^2 \sqrt {1+x^3} \, dx .$$

We compute the area of the part of the rectangle that is above the graph.  The function
$y=\sqrt {1+x^3}$ is monotonic on the segment $[0,2]$.  Thus, we can express $x$ in terms of
$y$:
$$x = \root 3 \of {y^2 - 1} .$$

Therefore, the area we seek is
$$\int_1^3 \root 3 \of {y^2 - 1} \, dy .$$
Make the following change of variable under the integral sign: $y=t+1$.  This yields the following
expression for the area:
$$\int_0^2 \root 3 \of {t^2 + 2t} \, dt .$$
Now we see that the given integral is equal to the area of the rectangle $OABC$, which is 6.

\ \\ \ \\

\noindent This problem is from \textit{How do you Figure? Math Challenges, M249},
Quantum, \textbf{9.2} (Nov/Dec 1998), 25, 45--46.
\ \\ \ \\
\noindent 5. Evaluate the series
$$\sum_{n=0}^\infty\frac1{2011^{2^n}-2011^{-{2^n}}}=\frac1{2011^1-2011^{-1}}
+\frac1{2011^2-2011^{-2}}+\frac1{2011^4-2011^{-4}}+\dots$$
and express it as a rational number.
\ \\ \ \\
\noindent Solution.
\ \\ \ \\
\noindent More generally, let
$$S(x)=\sum_{n=0}^\infty\frac{x^{2^n}}{1-x^{2^{n+1}}},\quad S_N(x)=\sum_{n=0}^N
\frac{x^{2^n}}{1-x^{2^{n+1}}}.$$
Note that the series in the problem is equal to $S(1/2011)$.  We will show that, for any $x$
with $0<x<1$,
\begin{equation}
S(x)=\frac x{1-x}.
\end{equation}
We use a ``telescoping'' argument based on the elementary identity
$$\frac{x^{2^n}}{1-x^{2^{n+1}}}=\frac1{1-x^{2^n}}-\frac1{1-x^{2^{n+1}}}.$$
If we substitute this into the partial sum $S_N(x)$, we get
$$S_N(x)=\bigg(\frac1{1-x}-\frac1{1-x^2}\bigg)+\bigg(\frac1{1-x^2}-\frac1{1-x^4}\bigg)+\dots
+\bigg(\frac1{1-x^{2^N}}-\frac1{1-x^{2^{N+1}}}\bigg)$$
which equals $\displaystyle\frac1{1-x}-\frac1{1-x^{2^{N+1}}}$.  As $N\rightarrow\infty$,
the second term approaches 1, thus
$$S(x)=\lim_{N\rightarrow\infty}S_N(x)=\frac1{1-x}-1=\frac x{1-x},$$
proving (2).

\noindent Therefore, the solution of the problem is $S(1/2011)=(1/2011)/(1-1/2011)=1/2010$.
\ \\ \ \\
\noindent (Similar to a U of I contest problem.)

\end{document}
